Question

Sin2α(альфа)=(sinα+cosα) "в квадрате" -1(1-cos2α)(1+cos2α)=sin2α ,буду

Answer 1

$sin2 \alpha =-1+cos^2 \alpha +cos^2 2\alpha + 2cos \alpha sin \alpha + sin^2 \alpha$
$1 - cos^2 \alpha -cos^22 \alpha - 2cos \alpha sin \alpha -sin^2 \alpha +sin2 \alpha = 0$
$\frac{1}{2} (-1-cos4 \alpha ) = 0$
$-1-cos4 \alpha = 0$
$-cos4 \alpha = 1$
$cos4 \alpha =- 1$
$4 \alpha = \pi +2 \pi n$
$\alpha = \frac{ \pi }{4} + \frac{ \pi n}{2}$ , $n$ ∈ $Z$